SOLUTION: 1/cube root of 4 + cube root of 5

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Question 719718: 1/cube root of 4 + cube root of 5
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, please
  • Include the instructions for the problem. Don't assume that there is only one obvious thing to do with an expression.
  • Put multiple-term numerators and denominators in parentheses. What you posted meant:
    1%2Froot%283%2C+4%29+%2B+root%283%2C+5%29
    But I'm pretty sure you intended:
    1%2F%28root%283%2C+4%29+%2B+root%283%2C+5%29%29
Assuming that the expression is:
1%2F%28root%283%2C+4%29+%2B+root%283%2C+5%29%29
and the instructions are "Rationalize the denominator" (or something to that effect), then we will take advantage of the factoring pattern:
%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29+=+a%5E3%2Bb%5E3
This pattern shows how a two-term expression times a certain three-term expression results in an expression of perfect cubes. The denominator we have to rationalize is a two-term expression which matches the pattern of (a+b). So the pattern shows how to turn that into an expression of perfect cubes. Since the cube of a cube root is rational, this is our path to a solution. In our denominator, the "a" is root%283%2C4%29 and the "b" is root%283%2C5%29. If we multiply the numerator and denominator by the corresponding a%5E2-ab%2Bb%5E2 we will reach our goal:

In the denominator the pattern tells us how it works out:

which simplifies to:
%28root%283%2C16%29-root%283%2C20%29%2Broot%283%2C+25%29%29%2F%284%2B5%29
then
%28root%283%2C16%29-root%283%2C20%29%2Broot%283%2C+25%29%29%2F9